Question: Simplify the following expression: $y = \dfrac{6x^2- 11x- 21}{x - 3}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(6)}{(-21)} &=& -126 \\ {a} + {b} &=& &=& {-11} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-126$ and add them together. Remember, since $-126$ is negative, one of the factors must be negative. The factors that add up to ${-11}$ will be your ${a}$ and ${b}$ When ${a}$ is ${7}$ and ${b}$ is ${-18}$ $ \begin{eqnarray} {ab} &=& ({7})({-18}) &=& -126 \\ {a} + {b} &=& {7} + {-18} &=& -11 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({6}x^2 +{7}x) + ({-18}x {-21}) $ Factor out the common factors: $ x(6x + 7) - 3(6x + 7)$ Now factor out $(6x + 7)$ $ (6x + 7)(x - 3)$ The original expression can therefore be written: $ \dfrac{(6x + 7)(x - 3)}{x - 3}$ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ This leaves us with $6x + 7; x \neq 3$.